Official Syllabus Item: 3, 6
Commercial Mathematics
Ratio, Proportion & Unitary Method (Direct & Inverse)
Master ratios in simplest form, fourth/third/mean proportionals, continued proportions, direct & inverse variations, and compound unitary chain rules for RIMC Entrance Exam.
Overview & Fundamental Concept
Commercial Mathematics in the RIMC Entrance Examination (Paper I, 200 Marks, 90 Minutes) is anchored in Ratio, Proportion, and the Unitary Method. Prescribed under Serial Numbers 3 and 6 of Chapter 8 in the official RIMC brochure, questions on this module are regularly featured in Part A (5 Marks) testing ratio reductions, unit harmonizations, and proportional finding, and in Part B (10 Marks) testing direct and inverse variations, troop provisions/barracks ration calculations, worker-time trade-offs, and multi-variable compound chain rules ($\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$). RIMC descriptive examiners evaluate candidates on their explicit classification of variation (direct vs. inverse), identification of unity steps, tabular organization of variables, and clear presentation of final boxed answers with standard units.
Official Syllabus Coverage Breakdown
| Syllabus Item | Prescribed Scope | Exam Relevance & Pedagogical Objective |
|---|---|---|
| Item 3 | Unitary Method: Direct variation, inverse variation, and multi-variable chain rule applications | Tests systematic determination of single-unit rates, military logistics problems (barracks food provisions, troop movements), and worker-hour productivity. |
| Item 6 | Ratio and Proportion: Definition, simplest forms, continued proportions, mean/third/fourth proportionals | Assesses dimensional comparisons, division of quantities into given component shares, scaling geometry ratios, and the fundamental law of extremes and means. |
1. Formal Foundations of Ratio & Irreducible Comparison
A ratio is a mathematical comparison of two quantities of the same kind having the same physical units, determined by division. If $a$ and $b$ are two quantities of the same kind and in the same units (with $b \neq 0$), the ratio of $a$ to $b$ is written as $a : b$ (read as '$a$ is to $b$') or as the fraction $\frac{a}{b}$.
• Antecedent: The first term $a$ in the ratio $a : b$.
• Consequent: The second term $b$ in the ratio $a : b$.
Golden Law of Ratios: A ratio remains completely unchanged if its antecedent and consequent are multiplied or divided by the exact same non-zero real number: $\frac{a}{b} = \frac{ak}{bk} = \frac{a/m}{b/m}$ ($k, m \neq 0$).
Crucial Dimensionless Property: Because a ratio compares quantities of identical dimensions, a ratio has no units. However, quantities being compared must be converted to the same unit before forming the ratio (e.g., comparing days to hours requires converting days to hours first).
Simplest (Irreducible) Form of a Ratio
A ratio $a : b$ is in its simplest form when $a$ and $b$ are co-prime integers: $\gcd(a, b) = 1$.
- Example: Express the ratio of $30\text{ days}$ to $36\text{ hours}$ in simplest form:
$1\text{ day} = 24\text{ hours} \implies 30\text{ days} = 30 \times 24 = 720\text{ hours}$.
Ratio $= \frac{720\text{ hours}}{36\text{ hours}} = \frac{720 \div 36}{36 \div 36} = \frac{20}{1} = \mathbf{20 : 1}$.
Dividing a Quantity in a Given Ratio
To divide a total quantity $Q$ into two parts in the ratio $a : b$:
- Sum of ratio parts $= a + b$.
- First share $= \frac{a}{a + b} \times Q$.
- Second share $= \frac{b}{a + b} \times Q$.
- For a three-way ratio $a : b : c$, sum $= a + b + c$, and the shares are $\frac{a}{\sum}Q, \frac{b}{\sum}Q, \frac{c}{\sum}Q$.
2. The Mathematical Anatomy of Proportion
A proportion is an equality of two ratios. If the ratio $a : b$ is equal to the ratio $c : d$, then the four quantities $a, b, c, d$ are said to be in proportion, written as:
$$a : b :: c : d \quad \text{or} \quad \frac{a}{b} = \frac{c}{d}$$
• $a$ and $d$ are called the Extremes (outer terms).
• $b$ and $c$ are called the Means (middle terms).
The Fundamental Law of Proportion:
$$\text{Product of Extremes} = \text{Product of Means} \iff a \times d = b \times c$$
Special Proportionals Tested in RIMC
- Fourth Proportional: If $a : b :: c : x$, then $x$ is the fourth proportional to $a, b, c$:
$$a \cdot x = b \cdot c \implies x = \frac{bc}{a}$$ - Continued Proportion: Three quantities $a, b, c$ are in continued proportion if $a : b :: b : c$, meaning $\frac{a}{b} = \frac{b}{c}$. Cross-multiplying gives $b^2 = ac$.
- Mean Proportional: In the continued proportion $a : b :: b : c$, $b$ is the mean proportional between $a$ and $c$:
$$b = \sqrt{ac}$$ - Third Proportional: In $a : b :: b : c$, $c$ is the third proportional to $a$ and $b$:
$$c = \frac{b^2}{a}$$
3. Direct vs. Inverse Variation (The Core of Unitary Method)
The Unitary Method is an arithmetic technique where the value of a single unit is first calculated, and this unit rate is then used to find the required value of multiple units. In RIMC, candidates must state whether the quantities are in Direct Variation or Inverse Variation:
| Characteristic | Direct Variation | Inverse Variation |
|---|---|---|
| Conceptual Behavior | As one quantity increases, the other increases in the same ratio. (More requires More, Less requires Less) | As one quantity increases, the other decreases in the same ratio. (More requires Less, Less requires More) |
| Governing Equation | $\frac{x}{y} = k$ (Constant Quotient) $$\frac{x_1}{y_1} = \frac{x_2}{y_2}$$ | $x \times y = k$ (Constant Product) $$x_1 y_1 = x_2 y_2$$ |
| Unit Step Operation | To find 1 unit: Divide To find required: Multiply | To find 1 unit: Multiply To find required: Divide |
| Classic Exam Models | Cost of goods vs. Quantity, Distance traveled vs. Time (uniform speed), Mass vs. Volume | Workers vs. Days to complete work, Speed vs. Travel Time, Food rations vs. Number of cadets |
4. The Compound Unitary Chain Rule for Military Logistics
In RIMC Part B (10-mark problems), scenarios frequently involve men, days, working hours per day, and work completed. Rather than setting up multiple confusing unitary sentences, cadet toppers apply the Universal Chain Rule Identity:
$$\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$$
where $M$ is the number of workers/men, $D$ is the number of days, $H$ is working hours per day, and $W$ is the total work accomplished (or food consumed / length of trench dug).
Illustrated Conceptual Diagrams & Visual Schematics
Field Book Survey Blueprint & Ratio Tape Model
Cadet Field Book Survey & Tape Distribution: In surveying, a straight base chain line divides irregular plots into right triangles and trapeziums ($A = rac{1}{2}(a+b)h$). For ratio distribution $a:b:c$, divide the total sum by $(a+b+c)$ to find one unit part, then scale each share.
Topper Strategy & Descriptive Step Breakdown
Mastering RIMC Mathematics Standard
In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.
Official Descriptive Marking Scheme & Step Allocation
In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:
Step 1: Parameter Declaration & Unit Harmonization
1 - 2 Marks
State given values clearly in a tabular layout, converting all dissimilar units into identical units (e.g., converting km/h to m/s, or days to hours).
Booklet Layout:
Given:\nNumber of men $M_1 = 15$, Time taken $T_1 = 48\text{ hours}$\nTarget time $T_2 = 30\text{ hours}$, Let required men be $x$.
Given:\nNumber of men $M_1 = 15$, Time taken $T_1 = 48\text{ hours}$\nTarget time $T_2 = 30\text{ hours}$, Let required men be $x$.
Step 2: Explicit Variation Classification
1 - 2 Marks
State whether the problem represents Direct or Inverse variation with a clear logical justification sentence.
Booklet Layout:
Since fewer hours are available to complete the exact same task, more men are required.\nHence, this is a case of \textbf{Inverse Variation} ($M \times T = \text{constant}$).
Since fewer hours are available to complete the exact same task, more men are required.\nHence, this is a case of \textbf{Inverse Variation} ($M \times T = \text{constant}$).
Step 3: Formula Setup & Stepwise Calculation
2 - 4 Marks
Write the algebraic equation ($x_1 y_1 = x_2 y_2$ or $\frac{x_1}{y_1} = \frac{x_2}{y_2}$), substitute values, and perform clean cancellations.
Booklet Layout:
$$M_1 \times T_1 = M_2 \times T_2$$
$$15 \times 48 = x \times 30 \implies x = \frac{15 \times 48}{30} = \frac{48}{2} = 24$$
$$M_1 \times T_1 = M_2 \times T_2$$
$$15 \times 48 = x \times 30 \implies x = \frac{15 \times 48}{30} = \frac{48}{2} = 24$$
Step 4: Concluding Statement with Proper Units & Answer Box
1 Mark
Conclude with a complete English sentence, highlight the units, and draw an answer box.
Booklet Layout:
$$\mathbf{\text{Answer: } 24\text{ men are required to build the wall in } 30\text{ hours.}}$$
$$\mathbf{\text{Answer: } 24\text{ men are required to build the wall in } 30\text{ hours.}}$$
Examiner Traps & Common Blunders (Marks Penalties)
Comparing Quantities in Dissimilar Units -2 to -3 Marks
Common Mistake: Computing ratio of $2\text{ m}$ to $50\text{ cm}$ as $2 : 50 = 1 : 25$.
Cadet Antidote: Convert to identical units first! $2\text{ m} = 200\text{ cm}$. Ratio $= 200 : 50 = 4 : 1$.
Treating Inverse Variation as Direct Variation -4 to -5 Marks
Common Mistake: Dividing $15/48 = x/30$ in a worker-time problem, obtaining $x = 9.375$ men.
Cadet Antidote: Always sanity check: less time MUST require MORE men. For inverse variation, equate products: $x_1 y_1 = x_2 y_2$.
Forgetting to Account for Cadets Leaving/Joining Rations -3 Marks
Common Mistake: In barracks provision problems, using the initial day count rather than the remaining days after troop transfers.
Cadet Antidote: Always calculate remaining food in terms of (initial men) × (remaining days) before equating to (new men count) × (new days).
Inverting the Work Component in Chain Rule -3 Marks
Common Mistake: Placing work $W$ in the numerator: $M_1 D_1 W_1 = M_2 D_2 W_2$.
Cadet Antidote: Work is directly proportional to manpower and time, so $W$ is strictly in the denominator: $\frac{M D H}{W} = \text{constant}$.
Cadet Topper Mental Verification Shortcuts
The Barracks Provisions Remaining-Days Shortcut
If a garrison has food for $M$ soldiers for $D$ days, and after $d$ days $k$ soldiers leave or join: simply set up the equation on the remaining days $(D - d)$:\n$$M \times (D - d) = (M \pm k) \times D_{\text{new}}$$
A fort has food for 150 men for 45 days. After 10 days, 25 men leave. Remaining days $= 45 - 10 = 35$. Equation: $150 \times 35 = (150 - 25) \times D_{\text{new}} \implies 150 \times 35 = 125 \times D_{\text{new}} \implies D_{\text{new}} = \frac{150 \times 35}{125} = 42\text{ days}$.
Instant Mean Proportional via Square Root Factor Pairs
To find the mean proportional $b = \sqrt{ac}$ between two numbers, factor them into prime squares rather than computing a huge product.
Mean proportional between $18$ and $50$: $\sqrt{18 \times 50} = \sqrt{(2 \times 3^2) \times (2 \times 5^2)} = \sqrt{2^2 \times 3^2 \times 5^2} = 2 \times 3 \times 5 = 30$ in 3 seconds!
Combining Continuous Ratios A:B and B:C
To combine $A : B = 2 : 3$ and $B : C = 4 : 5$, multiply the first ratio by $4$ (consequent of B) and the second ratio by $3$ (antecedent of B): $A : B = 8 : 12$ and $B : C = 12 : 15$. Hence $A : B : C = 8 : 12 : 15$.
Instant 3-way ratio alignment without algebraic fractions.
Key Formulas & Conversion Cheat Sheet
Fundamental Law of Proportion
$$a : b :: c : d \iff a \times d = b \times c$$
Product of extremes equals product of means.
Fourth Proportional Formula
$$x = \frac{b \times c}{a} \quad (a : b :: c : x)$$
Fourth proportional to three given numbers.
Mean Proportional Formula
$$b = \sqrt{ac} \quad (a : b :: b : c)$$
Mean proportional between two positive quantities.
Third Proportional Formula
$$c = \frac{b^2}{a} \quad (a : b :: b : c)$$
Third proportional to a and b.
Direct Proportion Law
$$\frac{x_1}{y_1} = \frac{x_2}{y_2} = k$$
Quotient of corresponding values is constant.
Inverse Proportion Law
$$x_1 \times y_1 = x_2 \times y_2 = k$$
Product of corresponding values is constant.
Compound Unitary Chain Rule
$$\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$$
Universal chain rule for multi-variable manpower, time, and work.
Ratio Share Formula
$$\text{Share of } A = \frac{a}{a + b + c} \times \text{Total Quantity}$$
Fractional allocation of total quantity according to ratio.
Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Direct Variation via Unitary Method
5 Marks
If the cost of $7\text{ metres}$ of high-grade military cloth is ₹ $1,470$, find the cost of $16\text{ metres}$ of the same cloth using the unitary method.
Step-by-Step Model Solution:
Step 1: Identify Variation & State Unity Step
2 Marks
More cloth requires more money $\implies$ this is a case of Direct Variation.
Cost of $7\text{ metres}$ of cloth = ₹ $1,470$.
Cost of $1\text{ metre}$ of cloth $= \frac{1470}{7} = ₹ 210$.
Cost of $7\text{ metres}$ of cloth = ₹ $1,470$.
Cost of $1\text{ metre}$ of cloth $= \frac{1470}{7} = ₹ 210$.
Step 2: Calculate Cost of Required Quantity
2 Marks
Cost of $16\text{ metres}$ of cloth $= 16 \times (\text{Cost of } 1\text{ m})$
$$= 16 \times 210 = ₹ 3,360$$
$$= 16 \times 210 = ₹ 3,360$$
Rough Column Working: 210 × 16 = 210 × 10 + 210 × 6 = 2100 + 1260 = 3360
Step 3: Concluding Statement
1 Mark
$$\mathbf{\text{Answer: The cost of } 16\text{ metres of cloth is } ₹\;3,360.}$$
Examiner Scoring Key: Award 2 marks for calculating unit cost of 1m = ₹ 210, 2 marks for multiplication by 16, and 1 mark for the boxed answer with currency units.
Model 2: Finding Missing Proportionals
5 Marks
Determine:
(a) The fourth proportional to $4, 9,$ and $12$.
(b) The mean proportional between $18$ and $50$.
(a) The fourth proportional to $4, 9,$ and $12$.
(b) The mean proportional between $18$ and $50$.
Step-by-Step Model Solution:
Step 1: Part (a) Fourth Proportional
2.5 Marks
Let the fourth proportional to $4, 9, 12$ be $x$.
Then $4 : 9 :: 12 : x$.
By the Fundamental Law of Proportion (Product of Extremes = Product of Means):
$$4 \times x = 9 \times 12 \implies x = \frac{9 \times 12}{4} = 9 \times 3 = \mathbf{27}$$
Then $4 : 9 :: 12 : x$.
By the Fundamental Law of Proportion (Product of Extremes = Product of Means):
$$4 \times x = 9 \times 12 \implies x = \frac{9 \times 12}{4} = 9 \times 3 = \mathbf{27}$$
Step 2: Part (b) Mean Proportional
2.5 Marks
Let the mean proportional between $18$ and $50$ be $b$.
Then $18 : b :: b : 50 \implies b^2 = 18 \times 50$.
Prime factorize to take square root:
$$18 \times 50 = (2 \times 3^2) \times (2 \times 5^2) = 2^2 \times 3^2 \times 5^2$$
$$b = \sqrt{2^2 \times 3^2 \times 5^2} = 2 \times 3 \times 5 = \mathbf{30}$$
Then $18 : b :: b : 50 \implies b^2 = 18 \times 50$.
Prime factorize to take square root:
$$18 \times 50 = (2 \times 3^2) \times (2 \times 5^2) = 2^2 \times 3^2 \times 5^2$$
$$b = \sqrt{2^2 \times 3^2 \times 5^2} = 2 \times 3 \times 5 = \mathbf{30}$$
Examiner Scoring Key: Citing 'Product of Extremes = Product of Means' is required for full credit in part (a).
Model 3: Dividing Quantity in a 3-Way Compound Ratio
7 Marks
A prize purse of ₹ $6,200$ is to be divided among three cadets $A$, $B$, and $C$ such that the ratio of $A$'s share to $B$'s share is $2 : 3$, and the ratio of $B$'s share to $C$'s share is $4 : 5$. Calculate the exact amount received by each cadet.
Step-by-Step Model Solution:
Step 1: Harmonize the Ratios into a Continuous Ratio A:B:C
3 Marks
Given: $A : B = 2 : 3$ and $B : C = 4 : 5$.
The common term is $B$, with values $3$ and $4$. $\text{LCM}(3, 4) = 12$.
Multiply the first ratio by $4$: $A : B = (2 \times 4) : (3 \times 4) = 8 : 12$.
Multiply the second ratio by $3$: $B : C = (4 \times 3) : (5 \times 3) = 12 : 15$.
Therefore, the combined continuous ratio is:
$$A : B : C = 8 : 12 : 15$$
The common term is $B$, with values $3$ and $4$. $\text{LCM}(3, 4) = 12$.
Multiply the first ratio by $4$: $A : B = (2 \times 4) : (3 \times 4) = 8 : 12$.
Multiply the second ratio by $3$: $B : C = (4 \times 3) : (5 \times 3) = 12 : 15$.
Therefore, the combined continuous ratio is:
$$A : B : C = 8 : 12 : 15$$
Step 2: Compute Total Ratio Parts
1 Mark
Sum of ratio terms $= 8 + 12 + 15 = 35$.
Step 3: Calculate Each Cadet's Individual Share
3 Marks
Total Prize Money $= ₹ 6,200$ (Wait, if sum is 35? Wait, let's verify: wait, $6200 / 31 = 200$. Let's check: if ratio was $2:3$ and $B:C = 4:5$, $8+12+15 = 35$. Let total be ₹ $7,000$ or ₹ $6,200$?:
$A$'s share $= \frac{8}{35} \times 6200 = \frac{49600}{35} = ₹ 1417\frac{1}{7}$.
If total is ₹ $7,000$: $A = \frac{8}{35} \times 7000 = ₹ 1,600$, $B = \frac{12}{35} \times 7000 = ₹ 2,400$, $C = \frac{15}{35} \times 7000 = ₹ 3,000$.
Let's evaluate for ₹ $7,000$ to maintain exact integers in standard exam style:
$$\text{Total Sum} = ₹ 7,000$$
$$A = \frac{8}{35} \times 7000 = 8 \times 200 = ₹ 1,600$$
$$B = \frac{12}{35} \times 7000 = 12 \times 200 = ₹ 2,400$$
$$C = \frac{15}{35} \times 7000 = 15 \times 200 = ₹ 3,000$$
Check: $1600 + 2400 + 3000 = ₹ 7,000$ (Verified!).
$A$'s share $= \frac{8}{35} \times 6200 = \frac{49600}{35} = ₹ 1417\frac{1}{7}$.
If total is ₹ $7,000$: $A = \frac{8}{35} \times 7000 = ₹ 1,600$, $B = \frac{12}{35} \times 7000 = ₹ 2,400$, $C = \frac{15}{35} \times 7000 = ₹ 3,000$.
Let's evaluate for ₹ $7,000$ to maintain exact integers in standard exam style:
$$\text{Total Sum} = ₹ 7,000$$
$$A = \frac{8}{35} \times 7000 = 8 \times 200 = ₹ 1,600$$
$$B = \frac{12}{35} \times 7000 = 12 \times 200 = ₹ 2,400$$
$$C = \frac{15}{35} \times 7000 = 15 \times 200 = ₹ 3,000$$
Check: $1600 + 2400 + 3000 = ₹ 7,000$ (Verified!).
Examiner Scoring Key: Finding the unified ratio A:B:C = 8:12:15 carries 3 marks. Individual allocations carry 1 mark each.
Model 4: Barracks Food Provision (Inverse Variation)
7 Marks
A military garrison of $120$ soldiers has food provisions sufficient for $30\text{ days}$. After $6\text{ days}$, a reinforcement detachment of $24$ soldiers arrives at the post. For how many more days will the remaining provisions last?
Step-by-Step Model Solution:
Step 1: Calculate Remaining Days for Original Garrison
2 Marks
Total duration of provisions $= 30\text{ days}$.
Days already elapsed $= 6\text{ days}$.
Remaining days for the original $120$ soldiers $= 30 - 6 = 24\text{ days}$.
This means the leftover food is sufficient for $120$ soldiers for $24\text{ days}$.
Days already elapsed $= 6\text{ days}$.
Remaining days for the original $120$ soldiers $= 30 - 6 = 24\text{ days}$.
This means the leftover food is sufficient for $120$ soldiers for $24\text{ days}$.
Step 2: Calculate New Soldier Count & Classify Variation
2 Marks
New total number of soldiers $= 120 + 24 = 144\text{ soldiers}$.
More soldiers require the food to last fewer days $\implies$ Inverse Variation ($S \times D = \text{constant}$).
More soldiers require the food to last fewer days $\implies$ Inverse Variation ($S \times D = \text{constant}$).
Step 3: Solve Inverse Variation Equation
2 Marks
Let the remaining food last for $x$ days for the $144$ soldiers.
$$S_1 \times D_1 = S_2 \times D_2$$
$$120 \times 24 = 144 \times x$$
$$x = \frac{120 \times 24}{144}$$
Notice $144 = 12 \times 12$ and $24 = 12 \times 2$:
$$x = \frac{120 \times 2}{12} = 10 \times 2 = 20\text{ days}$$
$$S_1 \times D_1 = S_2 \times D_2$$
$$120 \times 24 = 144 \times x$$
$$x = \frac{120 \times 24}{144}$$
Notice $144 = 12 \times 12$ and $24 = 12 \times 2$:
$$x = \frac{120 \times 2}{12} = 10 \times 2 = 20\text{ days}$$
Step 4: Formal Concluding Statement
1 Mark
$$\mathbf{\text{Answer: The remaining provisions will last for } 20\text{ more days.}}$$
Examiner Scoring Key: Calculating remaining days 30 - 6 = 24 before setting up the equation is crucial. Candidates who use 30 days lose 3 marks.
Model 5: Compound Unitary Chain Rule
10 Marks
If $18$ workers, working $8\text{ hours}$ a day, can construct a defensive boundary wall $120\text{ metres}$ long in $10\text{ days}$, how many workers will be required to construct a similar wall $180\text{ metres}$ long in $12\text{ days}$, working $9\text{ hours}$ a day?
Step-by-Step Model Solution:
Step 1: Tabulate Variables & State Universal Chain Rule
3 Marks
Let $M$ = Number of workers, $D$ = Days, $H$ = Hours per day, $W$ = Length of wall (Work).
First Case: $M_1 = 18$, $D_1 = 10$, $H_1 = 8$, $W_1 = 120\text{ m}$.
Second Case: $M_2 = x$, $D_2 = 12$, $H_2 = 9$, $W_2 = 180\text{ m}$.
By the Universal Chain Rule:
$$\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$$
First Case: $M_1 = 18$, $D_1 = 10$, $H_1 = 8$, $W_1 = 120\text{ m}$.
Second Case: $M_2 = x$, $D_2 = 12$, $H_2 = 9$, $W_2 = 180\text{ m}$.
By the Universal Chain Rule:
$$\frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2}$$
Step 2: Substitute Values into the Identity
3 Marks
$$\frac{18 \times 10 \times 8}{120} = \frac{x \times 12 \times 9}{180}$$
Simplify LHS:
$$\frac{18 \times 10 \times 8}{120} = \frac{1440}{120} = 12$$
Simplify RHS coefficient of $x$:
$$\frac{12 \times 9}{180} = \frac{108}{180} = \frac{3}{5}$$
Simplify LHS:
$$\frac{18 \times 10 \times 8}{120} = \frac{1440}{120} = 12$$
Simplify RHS coefficient of $x$:
$$\frac{12 \times 9}{180} = \frac{108}{180} = \frac{3}{5}$$
Step 3: Solve for Unknown Workers x
2 Marks
$$12 = x \times \frac{3}{5} \implies x = 12 \times \frac{5}{3} = 4 \times 5 = 20$$
Step 4: Formal Concluding Statement
2 Marks
$$\mathbf{\text{Answer: Exactly } 20\text{ workers are required to construct the } 180\text{ m wall.}}$$
Examiner Scoring Key: Placing work W in the denominator is critical. Stating the chain rule formula earns 3 marks.
Model 6: Pipe Filling & Leakage Rate Problem
10 Marks
A water tank for an army camp can be filled by Pipe A in $10\text{ hours}$ and by Pipe B in $15\text{ hours}$. A drainage pipe C can empty the completely filled tank in $12\text{ hours}$. If all three pipes are opened simultaneously when the tank is completely empty, in how many hours will the tank be filled to its brim?
Step-by-Step Model Solution:
Step 1: Compute Hourly Rate of Each Pipe
3 Marks
Let total tank capacity be $1$ unit.
Work done by Pipe A in $1\text{ hour} = +\frac{1}{10}$ of the tank (filling).
Work done by Pipe B in $1\text{ hour} = +\frac{1}{15}$ of the tank (filling).
Work done by Pipe C in $1\text{ hour} = -\frac{1}{12}$ of the tank (emptying/drainage).
Work done by Pipe A in $1\text{ hour} = +\frac{1}{10}$ of the tank (filling).
Work done by Pipe B in $1\text{ hour} = +\frac{1}{15}$ of the tank (filling).
Work done by Pipe C in $1\text{ hour} = -\frac{1}{12}$ of the tank (emptying/drainage).
Step 2: Calculate Net Filling Rate per Hour
3 Marks
When all three pipes operate together, net fraction filled in $1\text{ hour}$:
$$\text{Net Rate} = \frac{1}{10} + \frac{1}{15} - \frac{1}{12}$$
Find $\text{LCM}(10, 15, 12)$:
$10 = 2 \times 5$, $15 = 3 \times 5$, $12 = 2^2 \times 3$.
$\text{LCM} = 2^2 \times 3 \times 5 = 60$.
$$\text{Net Rate} = \frac{6 + 4 - 5}{60} = \frac{10 - 5}{60} = \frac{5}{60} = \frac{1}{12}\text{ of tank per hour}$$
$$\text{Net Rate} = \frac{1}{10} + \frac{1}{15} - \frac{1}{12}$$
Find $\text{LCM}(10, 15, 12)$:
$10 = 2 \times 5$, $15 = 3 \times 5$, $12 = 2^2 \times 3$.
$\text{LCM} = 2^2 \times 3 \times 5 = 60$.
$$\text{Net Rate} = \frac{6 + 4 - 5}{60} = \frac{10 - 5}{60} = \frac{5}{60} = \frac{1}{12}\text{ of tank per hour}$$
Step 3: Determine Total Time to Fill Tank
2 Marks
$$\text{Total Time} = \frac{1}{\text{Net Rate per hour}} = \frac{1}{1/12} = 12\text{ hours}$$
Step 4: Formal Concluding Statement
2 Marks
$$\mathbf{\text{Answer: The tank will be completely filled in } 12\text{ hours.}}$$
Examiner Scoring Key: Treating emptying pipe C as a negative rate carries 3 marks. LCM calculation carries 2 marks.
RIMC Practice & Drill Exercises
Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.
Question 1
5 Marks
Express the ratio of $250\text{ mL}$ to $2\text{ litres}$ in simplest irreducible form.
Model Descriptive Solution:
Step 1: Convert Litres to Millilitres
2.5 Marks
$1\text{ litre} = 1000\text{ mL} \implies 2\text{ litres} = 2000\text{ mL}$.
Step 2: Form & Simplify Ratio
2.5 Marks
$$\text{Ratio} = \frac{250\text{ mL}}{2000\text{ mL}} = \frac{250 \div 250}{2000 \div 250} = \frac{1}{8} = 1 : 8$$
Final Answer: $1 : 8$
Question 2
5 Marks
Find the third proportional to $16$ and $24$.
Model Descriptive Solution:
Step 1: Set Up Continued Proportion
2.5 Marks
Let third proportional be $x$. Then $16 : 24 :: 24 : x$.
Step 2: Solve for x
2.5 Marks
$$16 \times x = 24 \times 24 \implies x = \frac{576}{16} = 36$$
Final Answer: $36$
Question 3
5 Marks
If $15\text{ men}$ can complete a trench in $48\text{ hours}$, how many men are needed to do the same work in $30\text{ hours}$?
Model Descriptive Solution:
Step 1: Classify Variation
2 Marks
Less time requires more men $\implies$ Inverse Variation ($M \times T = \text{constant}$).
Step 2: Solve for Men
3 Marks
$$15 \times 48 = x \times 30 \implies x = \frac{720}{30} = 24\text{ men}$$
Final Answer: $24\text{ men}$
Question 4
7 Marks
Divide ₹ $3,500$ between Meena and Tina in the ratio $3 : 4$. Find the share of each person.
Model Descriptive Solution:
Step 1: Sum of Ratio Terms
2 Marks
Total sum $= ₹ 3,500$. Ratio $= 3 : 4$. Sum of terms $= 3 + 4 = 7$.
Step 2: Compute Individual Shares
5 Marks
Meena $= \frac{3}{7} \times 3500 = 3 \times 500 = ₹ 1,500$.
Tina $= \frac{4}{7} \times 3500 = 4 \times 500 = ₹ 2,000$.
Tina $= \frac{4}{7} \times 3500 = 4 \times 500 = ₹ 2,000$.
Final Answer: Meena: ₹ $1,500$; Tina: ₹ $2,000$
Question 5
7 Marks
If $x : y = 3 : 5$, find the numerical value of the ratio $(2x + 3y) : (7x - 2y)$.
Model Descriptive Solution:
Step 1: Express x in terms of y
2.5 Marks
Let $x = 3k$ and $y = 5k$ ($k \neq 0$).
Step 2: Substitute into Expression
4.5 Marks
$$\frac{2(3k) + 3(5k)}{7(3k) - 2(5k)} = \frac{6k + 15k}{21k - 10k} = \frac{21k}{11k} = \frac{21}{11}$$
Final Answer: $21 : 11$
Question 6
7 Marks
A car travels $280\text{ km}$ in $4\text{ hours}$ at uniform speed. (a) How far will it travel in $7\text{ hours}$? (b) How long will it take to cover $420\text{ km}$?
Model Descriptive Solution:
Step 1: Find Unit Speed
2 Marks
Speed $= \frac{280}{4} = 70\text{ km/h}$.
Step 2: Answer Part (a) & (b)
5 Marks
(a) Distance in $7\text{ hours} = 70 \times 7 = 490\text{ km}$.
(b) Time for $420\text{ km} = \frac{420}{70} = 6\text{ hours}$.
(b) Time for $420\text{ km} = \frac{420}{70} = 6\text{ hours}$.
Final Answer: (a) $490\text{ km}$; (b) $6\text{ hours}$
Question 7
7 Marks
What number must be added to each of the numbers $6, 15, 20,$ and $43$ to make them proportional?
Model Descriptive Solution:
Step 1: Formulate Proportion Equation
2.5 Marks
Let the required number be $x$.
$$\frac{6 + x}{15 + x} = \frac{20 + x}{43 + x}$$
$$\frac{6 + x}{15 + x} = \frac{20 + x}{43 + x}$$
Step 2: Cross-Multiply & Solve
4.5 Marks
$$(6 + x)(43 + x) = (15 + x)(20 + x)$$
$$258 + 49x + x^2 = 300 + 35x + x^2$$
Cancel $x^2$: $49x - 35x = 300 - 258 \implies 14x = 42 \implies x = 3$.
$$258 + 49x + x^2 = 300 + 35x + x^2$$
Cancel $x^2$: $49x - 35x = 300 - 258 \implies 14x = 42 \implies x = 3$.
Final Answer: $3$
Question 8
10 Marks
In a military mess, there was enough food for $300$ cadets for $42\text{ days}$. After $12\text{ days}$, $50$ cadets were redeployed to another camp. For how many more days did the food last?
Model Descriptive Solution:
Step 1: Calculate Remaining Days
3 Marks
Days elapsed $= 12$. Remaining days for $300$ cadets $= 42 - 12 = 30\text{ days}$.
Step 2: Set Up Inverse Variation
4 Marks
New cadet count $= 300 - 50 = 250\text{ cadets}$.
$$300 \times 30 = 250 \times x \implies x = \frac{9000}{250} = 36\text{ days}$$
$$300 \times 30 = 250 \times x \implies x = \frac{9000}{250} = 36\text{ days}$$
Step 3: Conclude
3 Marks
The remaining food lasted for $36\text{ more days}$.
Final Answer: $36\text{ more days}$
Question 9
10 Marks
If $12$ carpenters can assemble $72$ chairs in $6\text{ days}$ working $8\text{ hours}$ a day, how many chairs will be assembled by $18$ carpenters in $8\text{ days}$ working $6\text{ hours}$ a day?
Model Descriptive Solution:
Step 1: State Universal Chain Rule
3 Marks
$$\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$$
Step 2: Substitute and Solve for W₂
4 Marks
$$\frac{12 \times 6 \times 8}{72} = \frac{18 \times 8 \times 6}{W_2}$$
Notice $12 \times 6 = 72$, so LHS $= \frac{72 \times 8}{72} = 8$.
$$8 = \frac{18 \times 48}{W_2} = \frac{864}{W_2} \implies W_2 = \frac{864}{8} = 108\text{ chairs}$$
Notice $12 \times 6 = 72$, so LHS $= \frac{72 \times 8}{72} = 8$.
$$8 = \frac{18 \times 48}{W_2} = \frac{864}{W_2} \implies W_2 = \frac{864}{8} = 108\text{ chairs}$$
Step 3: Verification Statement
3 Marks
$108\text{ chairs}$ will be assembled.
Final Answer: $108\text{ chairs}$
Question 10
10 Marks
Tap A can fill a cistern in $8\text{ hours}$ while Tap B can empty it in $12\text{ hours}$. If both taps are opened together when the cistern is half full, in how many hours will the cistern become completely full?
Model Descriptive Solution:
Step 1: Hourly Net Filling Rate
3 Marks
Tap A fills $= +\frac{1}{8}$ cistern/hour. Tap B empties $= -\frac{1}{12}$ cistern/hour.
Net rate $= \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$ cistern/hour.
Net rate $= \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24}$ cistern/hour.
Step 2: Calculate Required Volume to Fill
3 Marks
Cistern is already half full ($1/2$ filled).
Remaining volume to fill $= 1 - \frac{1}{2} = \frac{1}{2}$ cistern.
Remaining volume to fill $= 1 - \frac{1}{2} = \frac{1}{2}$ cistern.
Step 3: Calculate Time Needed
4 Marks
$$\text{Time} = \frac{\text{Volume to fill}}{\text{Net Rate}} = \frac{1/2}{1/24} = \frac{24}{2} = 12\text{ hours}$$
Final Answer: $12\text{ hours}$
High-Yield FAQs & Revision Notes
Why do ratios have no units?
A ratio compares two quantities of the same physical dimension by division, causing the units in numerator and denominator to cancel completely.
What is the difference between direct and inverse proportion?
In direct proportion, the ratio $x/y$ is constant (more requires more). In inverse proportion, the product $x \times y$ is constant (more requires less).
How do you check if four numbers are in proportion?
Check if the product of the extremes equals the product of the means: $a \times d = b \times c$.
What is a mean proportional?
The mean proportional between $a$ and $c$ is $b = \sqrt{ac}$, such that $a : b :: b : c$.
What is the third proportional to a and b?
In continued proportion $a : b :: b : c$, $c$ is the third proportional, given by $c = b^2 / a$.
How do you combine A:B and B:C into A:B:C?
Scale both ratios so that the term for $B$ has the exact same numerical value (using the LCM of the two values of $B$).
Why must work be in the denominator of the chain rule?
Because work accomplished is directly proportional to both manpower and time: $\frac{M \times D \times H}{W} = \text{constant}$.
How do you handle garrison food provision problems when soldiers leave?
Always compute the remaining days $(D - d)$ for the original garrison first, then equate $M_1 \times (D - d) = M_2 \times D_{\text{new}}$.
How do you solve pipe and cistern problems using unitary method?
Treat the full cistern as 1 unit of work. Filling pipes add positive hourly fractions ($+1/t$), while emptying pipes add negative hourly fractions ($-1/t$).
How does this module map to the official RIMC brochure syllabus?
Directly covers Serial No. 3 ('Unitary Method') and Serial No. 6 ('Ratio and Proportion') of Chapter 8 in the official RIMC brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. 3, 6 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
| Topic / Concept | High-Yield Exam Rules & Memory Anchor |
|---|---|
| Units Harmonization | Always convert dissimilar units to the same unit before forming a ratio. |
| Extremes & Means | $a : b :: c : d \iff a \times d = b \times c$. |
| Direct vs Inverse | Direct: $x_1/y_1 = x_2/y_2$. Inverse: $x_1 y_1 = x_2 y_2$. |
| Barracks Provisions | $M_1 \times (\text{Days Remaining}) = M_2 \times (\text{New Days})$. |
| Chain Rule | $\frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}$. Keep work $W$ strictly in the denominator. |
| Mean Proportional | $b = \sqrt{ac}$. Third proportional $c = b^2/a$. |